How can $\lambda$ in Poisson distribution be constant if derived from binomial distribution for $\lambda=np$ while $n$ goes to infinity? 
My Problem. Above is the derivation for Poisson from binomial which I quoted from somewhere. I understand this derivation mostly, but my struggle is that I do not understand why we can say $\lambda=np$, and $\lambda$ is constant in Poisson distribution.

My trial. My thought is that in a binomial distribution the probability of a success (i.e. for a certain event to happen) for each trial is constant, so $p$ is constant. Then deriving Poisson from binomial, we let $n$ approach infinity, and should not $\lambda$ go to infinity as a result as $\lambda=np$?

My comment. Any kind of idea or help would be appreciated. Thank you!
 A: You are getting confused because the proof is missing a few prior details. Also, it abuses notation by making you think only a single binomial distribution is involved above. This is not the case. Here is a more accurate rendition of what the proof is showing:
Fix $\lambda > 0$. And suppose $X_n \sim Binomial(n, p_n)$ is a sequence of binomial variables such that:

*

*each has a different success probability $p_n$, and

*they satisfy the very special condition $np_n = \lambda$ for all large enough $n$.

Then, the proof shows $\lim\limits_{n \to \infty} P(X_n = k) = P(X = k)$ where $X \sim Poisson(\lambda)$.
So the proof isn't really deriving the Poisson distribution from the Binomial distribution. That is, the proof isn't showing Poisson is some special case of the Binomial.
Rather, it is saying that a very special collection of binomial variables, satisfying a very special condition $np_n = \lambda$, starts behaving more and more like the Poisson distribution for large $n$.
A: Implicitly we assume at the very beginning that $\lambda$ is a constant. $p$ decreases as $n\to\infty$ so as to keep the product $\lambda=np$ constant (otherwise the expression for $P(X=k)$ would not remain a valid probability).
A: When $\lambda$ is called a "constant" in this context, that means that among the probabilities $\dfrac{\lambda^k e^{-\lambda}}{k!},$ the number $\lambda$ remains the same as $k$ varies within the set $\{0,1,2,3,\ldots\}.$
But also, within this proof, $\lambda=np$ does not change as $n$ and $p$ change. As $n$ gets bigger, $p = \lambda/n$ gets smaller and $\lambda$ remains the same.
